if a child pulls an old wagon up a steep incline on a sidewalk, what does a child's work do?

i say, it would increase, due to the fact that it is a steep hill and it takes more work to pull the wagon up???

The child's work is done aganst gravity, and results in an increase in the potential energy of the wagon. That increase equals the work done, neglecting friction.

ko

You're correct! When a child pulls an old wagon up a steep incline on a sidewalk, the child's work does increase. This is because the child is exerting a force against the gravitational force acting on the wagon to lift it higher up the incline.

To calculate the work done by the child, you can use the equation:

Work = Force x Distance x cos(theta)

In this case, the force being applied by the child is the force of gravity on the wagon. The distance is the distance the wagon is pulled up the incline, and theta is the angle between the direction of the force and the direction of the displacement. Since the wagon is being pulled straight up the incline, theta would be 0, and cos(0) equals 1.

Neglecting the effects of friction, the work done by the child will be equal to the increase in potential energy of the wagon. This means that the child's work is ultimately used to increase the potential energy of the wagon, allowing it to be at a higher position on the incline.